【数据库】数据库练习题答案（校正版）

作业链接：https://www.cnblogs.com/wupeiqi/articles/5729934.html

• 注释语句为语句前添加 --

• 构建临时表 select * from (select * from ...) as B; 需要保证查询项目在临时表中存在

• join连表时，宾语是谁全部显示

1. 自行创建测试数据

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
--  Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `caption` varchar(32) NOT NULL,
  PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES (‘1‘, ‘三年二班‘), (‘2‘, ‘三年三班‘), (‘3‘, ‘一年二班‘), (‘4‘, ‘二年九班‘);
COMMIT;

-- ----------------------------
--  Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `cname` varchar(32) NOT NULL,
  `teacher_id` int(11) NOT NULL,
  PRIMARY KEY (`cid`),
  KEY `fk_course_teacher` (`teacher_id`),
  CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES (‘1‘, ‘生物‘, ‘1‘), (‘2‘, ‘物理‘, ‘2‘), (‘3‘, ‘体育‘, ‘3‘), (‘4‘, ‘美术‘, ‘2‘);
COMMIT;

-- ----------------------------
--  Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `student_id` int(11) NOT NULL,
  `course_id` int(11) NOT NULL,
  `num` int(11) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_score_student` (`student_id`),
  KEY `fk_score_course` (`course_id`),
  CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
  CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES (‘1‘, ‘1‘, ‘1‘, ‘10‘), (‘2‘, ‘1‘, ‘2‘, ‘9‘), (‘5‘, ‘1‘, ‘4‘, ‘66‘), (‘6‘, ‘2‘, ‘1‘, ‘8‘), (‘8‘, ‘2‘, ‘3‘, ‘68‘), (‘9‘, ‘2‘, ‘4‘, ‘99‘), (‘10‘, ‘3‘, ‘1‘, ‘77‘), (‘11‘, ‘3‘, ‘2‘, ‘66‘), (‘12‘, ‘3‘, ‘3‘, ‘87‘), (‘13‘, ‘3‘, ‘4‘, ‘99‘), (‘14‘, ‘4‘, ‘1‘, ‘79‘), (‘15‘, ‘4‘, ‘2‘, ‘11‘), (‘16‘, ‘4‘, ‘3‘, ‘67‘), (‘17‘, ‘4‘, ‘4‘, ‘100‘), (‘18‘, ‘5‘, ‘1‘, ‘79‘), (‘19‘, ‘5‘, ‘2‘, ‘11‘), (‘20‘, ‘5‘, ‘3‘, ‘67‘), (‘21‘, ‘5‘, ‘4‘, ‘100‘), (‘22‘, ‘6‘, ‘1‘, ‘9‘), (‘23‘, ‘6‘, ‘2‘, ‘100‘), (‘24‘, ‘6‘, ‘3‘, ‘67‘), (‘25‘, ‘6‘, ‘4‘, ‘100‘), (‘26‘, ‘7‘, ‘1‘, ‘9‘), (‘27‘, ‘7‘, ‘2‘, ‘100‘), (‘28‘, ‘7‘, ‘3‘, ‘67‘), (‘29‘, ‘7‘, ‘4‘, ‘88‘), (‘30‘, ‘8‘, ‘1‘, ‘9‘), (‘31‘, ‘8‘, ‘2‘, ‘100‘), (‘32‘, ‘8‘, ‘3‘, ‘67‘), (‘33‘, ‘8‘, ‘4‘, ‘88‘), (‘34‘, ‘9‘, ‘1‘, ‘91‘), (‘35‘, ‘9‘, ‘2‘, ‘88‘), (‘36‘, ‘9‘, ‘3‘, ‘67‘), (‘37‘, ‘9‘, ‘4‘, ‘22‘), (‘38‘, ‘10‘, ‘1‘, ‘90‘), (‘39‘, ‘10‘, ‘2‘, ‘77‘), (‘40‘, ‘10‘, ‘3‘, ‘43‘), (‘41‘, ‘10‘, ‘4‘, ‘87‘), (‘42‘, ‘11‘, ‘1‘, ‘90‘), (‘43‘, ‘11‘, ‘2‘, ‘77‘), (‘44‘, ‘11‘, ‘3‘, ‘43‘), (‘45‘, ‘11‘, ‘4‘, ‘87‘), (‘46‘, ‘12‘, ‘1‘, ‘90‘), (‘47‘, ‘12‘, ‘2‘, ‘77‘), (‘48‘, ‘12‘, ‘3‘, ‘43‘), (‘49‘, ‘12‘, ‘4‘, ‘87‘), (‘52‘, ‘13‘, ‘3‘, ‘87‘);
COMMIT;

-- ----------------------------
--  Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `gender` char(1) NOT NULL,
  `class_id` int(11) NOT NULL,
  `sname` varchar(32) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_class` (`class_id`),
  CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES (‘1‘, ‘男‘, ‘1‘, ‘理解‘), (‘2‘, ‘女‘, ‘1‘, ‘钢蛋‘), (‘3‘, ‘男‘, ‘1‘, ‘张三‘), (‘4‘, ‘男‘, ‘1‘, ‘张一‘), (‘5‘, ‘女‘, ‘1‘, ‘张二‘), (‘6‘, ‘男‘, ‘1‘, ‘张四‘), (‘7‘, ‘女‘, ‘2‘, ‘铁锤‘), (‘8‘, ‘男‘, ‘2‘, ‘李三‘), (‘9‘, ‘男‘, ‘2‘, ‘李一‘), (‘10‘, ‘女‘, ‘2‘, ‘李二‘), (‘11‘, ‘男‘, ‘2‘, ‘李四‘), (‘12‘, ‘女‘, ‘3‘, ‘如花‘), (‘13‘, ‘男‘, ‘3‘, ‘刘三‘), (‘14‘, ‘男‘, ‘3‘, ‘刘一‘), (‘15‘, ‘女‘, ‘3‘, ‘刘二‘), (‘16‘, ‘男‘, ‘3‘, ‘刘四‘);
COMMIT;

-- ----------------------------
--  Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `tid` int(11) NOT NULL AUTO_INCREMENT,
  `tname` varchar(32) NOT NULL,
  PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES (‘1‘, ‘张磊老师‘), (‘2‘, ‘李平老师‘), (‘3‘, ‘刘海燕老师‘), (‘4‘, ‘朱云海老师‘), (‘5‘, ‘李杰老师‘);
COMMIT;

SET FOREIGN_KEY_CHECKS = 1;

2. 查询“生物”课程比“物理”课程成绩高的所有学生的学号

    获取所有有生物课程的人（学号，成绩） - 临时表
    获取所有有物理课程的人（学号，成绩） - 临时表
    根据【学号】连接两个临时表：
        学号  物理成绩   生物成绩
        
    IF(Condition,A,B)意义:当Condition为TRUE时,返回A;当Condition为FALSE时,返回B

select A.student_id, sw, wl from
(select student_id, num as sw from score left join course on score.course_id = course.cid where course.cname="生物") as A
left join
(select student_id, num as wl from score left join course on score.course_id = course.cid where course.cname="物理") as B
on A.student_id=B.student_id where sw > if(isnull(wl), 0, wl);

3. 查询平均成绩大于60分的同学的学号和平均成绩

select student_id, avg(num) from score group by student_id having avg(num) > 60;

4. 查询所有同学的学号、姓名、选课数、总成绩

select score.student_id, stu.sname, count(score.course_id), sum(score.num)
    from
        score left join student stu on stu.sid = score.student_id
    group by student_id;

5. 查询姓“李”的老师的个数

select count(tid) from teacher where tname like "李%";

利用临时表查询，count(1)计数
select count(1) from (select tid from teacher where tname like "李%") as B;

6. 查询没学过“李平”老师课的同学的学号、姓名

思路：
	先查询李平的所有课ID，即cid；
	之后筛选出所有选过该课的学生id；
	之后在student中查询没有此id的学生， not in

select * from student where student.sid not in (
    select distinct student_id from score where score.course_id in
    (select course.cid from course left join teacher t on course.teacher_id = t.tid where t.tname = "李平老师")
    );

7. 查询学过“1”并且也学过编号“2”课程的同学的学号、姓名

思路：and 索引 必须是不同的列名
	筛选出学过上述课程1或课程2的同学
	根据student.id进行分组，二次筛选count(id)>1的同学即可

select B.student_id, student.sname from
    (select student_id from score where score.course_id = ‘1‘ or score.course_id = ‘2‘)as B
    left join student on student.sid = B.student_id group by B.student_id having count(B.student_id)>1;

8. 查询学过“李平”老师所有门课的同学的学号、姓名

思路：
	先查询出该老师所教课的id；
	之后查询成绩表中选中上述课的学生id；
	之后对其分组，并统计上过全部课的学生名单；
	其中，全部课的个数应重新查询。

select student_id from score where score.course_id in
    (select course.cid from course left join teacher on teacher.tid = course.teacher_id where teacher.tname = "李平老师")
    group by student_id
    having count(course_id) = (select count(course.cid) from course left join teacher on teacher.tid = course.teacher_id where teacher.tname = "李平老师");

9. 查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名

select A.student_id, kc1, kc2 from
    (select student_id, num as kc1 from score where course_id=‘1‘) as A
    inner join
    (select student_id, num as kc2 from score where course_id=‘2‘) as B
    on A.student_id=B.student_id where kc1 > kc2;

10. 查询有课程成绩小于60分的同学的学号、姓名

select sid, sname from
(select distinct student_id from score where num < 60) as A
left join student on A.student_id=student.sid;

11. 查询没有学全所有课的同学的学号、姓名

思路：
	在分数表中根据学生进行分组，获取每一个学生选课数量；
	如果数量 == 总课程数量，表示已经选择了所有课程。

select s.sid, sname from score
left join
student s on score.student_id = s.sid
group by student_id having count(course_id) = (select count(cid) from course);

12. 查询至少有一门课与学号为“1”的同学所学相同的同学的学号和姓名

思路：
	查询该同学总共学了哪几门课；
	获取课程在其中的所有人以及所有课程
    根据学生筛选，获取所有学生信息
    再与学生表连接，获取姓名

select sid, sname from
    (select distinct student_id from score where course_id in
        (select course_id from score where student_id=‘1‘)) as A
    left join student on A.student_id=student.sid where sid != ‘1‘;

select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id;

13.查询至少学过学号为“1”同学所有课的其他同学学号和姓名

• in 之后保留相同的，而剔除不相同的
• (1,2,3)，用 in (1,2,4)则只会筛选出(1,2)

提醒：查询时，若查询的项为(1,2,3)，用 in (1,2,4)则只会筛选出(1,2)
思路：
	筛选出该同学所有课程id；
	判断score表中in这些课的数据行；
	设定count(course_id)=3，是为了筛除(1,2,3)等情况，而不影响(1,2,3,4)

select student_id, sname,count(course_id)
from score left join student on score.student_id = student.sid
    where student_id != 1 and course_id in (select course_id from score where student_id=1)
    group by student_id having count(course_id) = (select count(1) from score where student_id=1);

14. 查询和“2”号的同学学习的课程完全相同的其他同学学号和姓名

思路：
	满足两个条件：1.与该同学学习课程数一样的同学筛选出来 and 2. 学习的课程种类相同
	提醒：条件2在利用in之后，只会留下相同的course，此时course数不一定等于2同学的，因此需要count

select student_id,sname from score
left join student on score.student_id = student.sid
where student_id in
(select student_id from score where student_id != 2 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1))
and course_id in (select course_id from score where student_id = 2) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1);

15. 查询学习“李平”老师课的score表记录

select * from score where course_id in
(select cid from course left join teacher t on course.teacher_id = t.tid where tname="李平老师");

--------------16. 向score表中插入符合条件的记录：①没有上过编号“2”课程的同学学号；②插入“2”号课程的平均成绩

---------------17. 按平均成绩从低到高显示所有学生的“生物”、“物理”、“体育”三门的课程成绩，按如下形式显示: 学生ID、生物、物理、体育、选修课程数、有效平均分

本题中要求，
1 90 85 99
2 85 65 96
即，同一同学的同一行同时显示这三门课程，需要用到上图的策略。
在sql逐行查询时，将该时刻查询的student_id存入s1，之后传到子查询语句中（括号内）

18. 查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分

select score.course_id, max(num) as max_num, min(num) as min_num 
from score group by course_id;

19. 按各科平均成绩从低到高和及格率的百分数从高到低顺序排列：课程代号、平均分、及格率

• if(判断条件，成立结果，不成立结果)
  

select course_id, avg(num),
sum(if(score.num > 60, 1, 0))/count(1) * 100 as percent  先计算sum，再计算除法
from score group by course_id;

20. 课程平均分从高到低显示：平均分、任课老师

select tname, avg(num) from score
left join course c on score.course_id = c.cid
left join teacher t on t.tid = c.teacher_id
group by course_id;

--------------------21. 查询各科成绩前三名的记录

22. 查询每门课程被选修的学生数

select course_id, count(1) as number from score group by course_id;

23. 查询只选修一门课程的全部学生的学号和姓名

select s.sid, sname from score
left join student s on s.sid = score.student_id
group by student_id having count(1)=1;

24. 查询男生、女生的人数

select gender, count(1) from student
group by gender;

25. 查询姓“张”的学生名单

select sname from student where sname like "张%";

26. 查询同名同姓学生名单，并统计同名人数

select sname, count(1) from student group by sname having count(1)>1;

27. 查询每门课程的平均成绩，结果按平均成绩升序排列，相同时按课程号降序排列

select course_id, avg(num) from score group by course_id order by avg(num), course_id desc;

28. 查询平均成绩大于85的所有学生的学号、姓名和平均成绩

select student_id as sid, sname, avg(num) from score
left join student s on s.sid = score.student_id
group by student_id having avg(num) > 85;

29. 查询课程名称为“生物”，且分数低于60的学生姓名和分数

select student_id, sname from score
left join course c on c.cid = score.course_id
left join student s on score.student_id = s.sid
where c.cname = "生物" and score.num < 60;

30. 查询课程编号为3且课程成绩在80分以上的学生的学号和姓名

select student_id as sid, sname from course
left join score s on course.cid = s.course_id
left join student s2 on s2.sid = s.student_id
where course_id=3 and s.num > 80;

31. 求选课的学生人数

select count(distinct student_id) from score;

select count(1) from (select count(distinct student_id) from score 
group by student_id) as A;

32. 查询选修“张磊”老师所授课程的学生中，单科成绩最高的学生姓名及其成绩

select student_id, sname from score
left join student s on s.sid = score.student_id
where course_id in(
select course.cid from course left join teacher t on t.tid = course.teacher_id where tname="张磊老师")
order by num desc limit 1;

33. 查询各个课程及相应的选修人数

select course_id, count(1) from score group by course_id;

* 34. 查询不同课程但成绩相同的学生的学号、课程号、学生成绩

两张表放一起，select * from score as s1, score as s2 笛卡尔积（行数*行数）

select s1.student_id from score as s1, score as s2
where s1.sid != s2.sid and s1.num=s2.num and s1.course_id != s2.course_id;

---s1.sid != s2.sid 删除相同记录的数据行

* 35. 查询每门课程成绩最好的前两名

36. 查询至少选修两门课程的学生学号

select student_id from score group by student_id having count(1) >1;

37. 查询全部学生都选修的课程的课程号和课程名

select course_id from score group by course_id having count(1) = (select count(1) from student);

38. 查询没学过“张磊”老师讲授的任一门课程的学生姓名

select student_id from score where student_id not in
(select student_id from score
left join course c on c.cid = score.course_id
left join teacher t on t.tid = c.teacher_id
where t.tname = "张磊老师");

39. 查询两门以上不及格课程的同学的学号及其平均成绩

select student_id, avg(num) from score
where num < 60 group by student_id having count(1) >2;

40. 检索“4”课程分数小于60，按分数降序排列的同学学号

select student_id from score where course_id=4 and num < 60 order by num desc;

41. 查询“2”同学的“1”课程的成绩

select num from score where student_id=2 and course_id=1;

【数据库】数据库练习题答案（校正版）

原文：https://www.cnblogs.com/holaplace/p/14004803.html